Question

$\int\limits^\frac{\pi}{2}_0 {\frac{cosx}{1+sin^{2}x} \, dx$

Answer 1

$\int\limits^{\frac{\pi}{2}}_0 {\frac{\cos x}{1+\sin^2 x}} \, dx =\left[\begin{array}{c}{\sin x=t}&{dt=d(\sin x)=(\sin x)\prime dx=\cos x dx}&{0\leq x\leq\frac{\pi}{2}\rightarrow 0\leq t\leq 1}\end{array}\right] =\\ \\ =\int\limits^1_0 {\frac{dt}{1+t^2}} \, =arctg t|\limits^1 _0=arctg1-arctg0=\frac{\pi}{4}-0= \frac{\pi}{4}$